What is the mechanics of the union-intersection along a sequence of sets?

Question

This question came about after I got stuck trying to understand this answer.

When $n\to \infty$ I don't understand why in a countable family of sets $(A_n)_{n\geq 0}$ the operation

$$\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $$

depends on whether the sequence is monotonous or not.

It is likely that I don't understand the symbols, or that I am just too dumb. Let's see... Should I read this as a composition of operations? If so, when I start from the inside, i.e.

$$\bigcup_{k \geq n} A_k$$

I necessarily end up with all the elements in all the sets combined into a single set. At that point this entire set containing all possible elements in any one of the sets $A_k$ is intersect-ioned with all of the sets $A_k,$ looking for elements that are in all sets. Is that it? If this is correct, then why does the order of the monotony of the sequence matter? Can I get an example?

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Post-mortem:

' For a finite collection of sets:

$$ \begin{align} A_0&=\{a,b\}\\ A_1&=\{a,x\}\\ A_3&=\{a,x,y\}\\ A_4&=\{a,c,d,y\}\\[3ex] \bigcap_{n\geq 0} \bigcup_{k \geq n} A_k &=\{a,b,x,y,c,d\}\cap\{a,x,y,c,d\}\cap\{a,x,y,c,d\}\\ &=\{a,x,y,c,d\}\\[3ex] \bigcup_{n\geq 0} \bigcap_{k \geq n} A_k &=\{a\}\cup\{a\}\cup\{a,y\}=\{a,y\} \end{align} $$

In pseudo-code $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $:

for $n$ in $n=0$ to $\infty:$

for $k$ in $n$ to $\infty:$

$B_k \leftarrow A_k \cup A_{k+1} \cup A_{k+2} \cdots$

$B_0 \cap B_1 \cap B_2 \cap \cdots$

The pictorial intuition is that for an element $\color{blue}\spadesuit$ to be in $\bigcap_{n\geq 0} \bigcup_{k \geq n} A_k $ it would have to keep on appearing so that no matter how far we go into infinity it appears in some subset to as to be part of all $B_k$'s, and hence, in the intersection:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

On the other hand, $\bigcup_{n\geq 0} \bigcap_{k \geq n} A_k $ is more demanding, requiring that $\color{blue}\spadesuit$ appears consistently after a certain point:

$$\{\cdots\},\cdots,\{\cdots\},\{\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit\},\{\color{blue}\spadesuit,\cdots\},\cdots,\{\color{blue}\spadesuit,\cdots\},\{\cdots,\color{blue}\spadesuit,\cdots\},\cdots$$

Answer 1

Let me suggest a way of thinking about

$$\bigcap_{n\ge 0}\bigcup_{k\ge n} A_k$$

and

$$\bigcup_{n\ge 0}\bigcap_{k\ge n} A_k$$

that may be helpful.

How does something get into either of these sets?

• $x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x\in\bigcup\limits_{k\ge n}A_k$ for each $n\ge 0$, which is the case if and only if for each $n\ge 0$ there is a $k\ge n$ such that $x\in A_k$. And this is the same as saying that $x$ is in infinitely many of the sets $A_k$: no matter how far out in the sequence of sets you go, there is one further out that contains $x$.
• $x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n}A_k$ if and only if there is an $n\ge 0$ such that $x\in\bigcap\limits_{k\ge n}A_k$, which is the case if and only if there is an $n\ge 0$ such that $x\in A_k$ for every $k\ge n$. This just says that $x$ is in every $A_k$ from some point on: it is in all but finitely many of the sets $A_k$.

To sum up:

• $x\in\bigcap\limits_{n\ge 0}\bigcup\limits_{k\ge n} A_k$ if and only if $x$ belongs to infinitely many of the sets $A_k$, and
• $x\in\bigcup\limits_{n\ge 0}\bigcap\limits_{k\ge n} A_k$ if and only if $x$ is in all but finitely many of the sets $A_k$ or, equivalently, $x$ is in every $A_k$ from some point on.

Answer 2

Read it from left to right, like a double summation. First, we choose $n$. Then $\bigcup_{k\geq n}A_k$ is the set $B_n$, say, of all elements that are in one of the $A_k$ with index $\geq n$. $$B_n = A_n\cup A_{n+1}\cup A_{n+2}\cup\cdots$$ and $\bigcap_{n\geq0}\bigcup_{k\geq n}A_k$ is the set of elements in all of the $B_n$ $$\bigcap_{n\geq0}\bigcup_{k\geq n}A_k = B_0\cap B_1\cap B_2\cap\cdots$$ The intersection is not redundant. Consider an element $x$ that is in $A_0$ but none of the other $A_n$. Then $x\in B_0, x\notin B_1,\ x\notin B_2, \dots$ so $x\notin C:=\bigcap_{n\geq0}\bigcup_{k\geq n}A_k$

In fact, C is precisely the set of elements in infinitely many of the $A_n$. (If $x$ is in only finitely many of the $A_n$ then there is some $n$ such that $n\geq N\implies x\notin A_n$. Consequently, $x\notin B_N$.)

You might want to convince yourself that if we interchange the union and intersection, then $$\bigcup_{n\geq0}\bigcap_{k\geq n}A_k$$ is the set of all elements is all but finitely many of the $A_n$

Answer 3

They are answering the specific question "wouldn't the $\bigcup_{k \geq n} A_k$ part just include everything, leaving the intersection part redundant?" The answer is no, only if the $A_k$ are monotone increasing to $\mathbb{R}$.

Your interpretation of $\bigcap_{n \geq 0} \bigcup_{k \geq n} A_k$ is exactly correct. The comments in the other post are just pointing out that $\bigcup_{k \geq n} A_k$ is not necessarily $\mathbb{R}$ (although it is in their example, which is perhaps why the comments are confusing).